Problem: The equation of a circle $C$ is $x^2+y^2-18x-8y+96 = 0$. What is its center $(h, k)$ and its radius $r$ ?
To find the equation in standard form, complete the square. $(x^2-18x) + (y^2-8y) = -96$ $(x^2-18x+81) + (y^2-8y+16) = -96 + 81 + 16$ $(x-9)^{2} + (y-4)^{2} = 1 = 1^2$ Thus, $(h, k) = (9, 4)$ and $r = 1$.